A wholesale distributor stocks and sells low flow toilets to contractors for use

Question

A wholesale distributor stocks and sells low flow toilets to contractors for use in commercial office buildings. The estimated annual demand for the toilets is 5,475 units. The estimated average demand per day is 18 units. The purchase cost from the toilet manufacturer is $135.00 per unit. The lead-time for a new order is 5 days. The ordering cost is $90.00 per order. The average holding cost per unit per year is $4.05. The distributor has traditionally ordered 210 units each time they placed an order. Based upon using the distributor’s current ordering model: (5 points each for a total of 25 points)

Task 5. The president of the wholesale distributor is concerned about the possibility of stockouts causing a loss of customer confidence and loyalty and is interested in maintaining safety stock in inventory to prevent potential stockouts. Based upon using the safety stock models: (10 points each for a total of 30 points)

: Assuming demand is normally distributed with a mean of 18 units and a standard deviation of 4 units, and a constant lead time of 5 days, what is the reorder point necessary to provide a 98% level of service?

: Assuming demand is constant at 18 units per day, and lead time is normally distributed with a mean of 5 days and a standard deviation of 2 days, what is the reorder point necessary to provide a 98% level of service?

Assuming that demand is normally distributed with a mean of 18 units and a standard deviation of 4 units, and lead-time is normally distributed with a mean of 5 days and a standard deviation of 2 days, what is the reorder point necessary to provide a 98% level of service?

Explanation / Answer

Question 1

Assuming demand is normally distributed with a mean of 18 units and a standard deviation of 4 units, and a constant lead time of 5 days, what is the reorder point necessary to provide a 98% level of service?

Solution:

We are given mean = 18, SD = 4 and n = 5, confidence level = 98%

We have to find 98% confidence interval for the mean demand.

The formula for confidence interval is given as below:

Confidence interval = Xbar -/+ t*SD/sqrt(n)

Degrees of freedom = n – 1 = 5 – 1 = 4

Critical value = t = 3.7469

Confidence interval = 18 -/+ 3.7469*4/sqrt(5)

Confidence interval = 18 -/+ 6.7027

Lower limit = 18 – 6.7027 = 11.30

Upper limit = 18 + 6.7027 = 24.70

Confidence interval = (11.30, 24.70)

Necessary recorder points = 11.30 and 24.70

Question 2

Assuming demand is constant at 18 units per day, and lead time is normally distributed with a mean of 5 days and a standard deviation of 2 days, what is the reorder point necessary to provide a 98% level of service?

Solution:

Mean = 5, SD = 2 and n = 18

Confidence interval = Xbar -/+ t*SD/sqrt(n)

Degrees of freedom = 18 – 1 = 17

Confidence level = 98%

Critical value = t = 2.5669

Confidence interval = 5 -/+ 2.5669*2/sqrt(18)

Lower limit = 5 - 2.5669*2/sqrt(18)

Lower limit = 3.79

Upper limit = 5 + 2.5669*2/sqrt(18)

Upper limit = 6.21

Confidence interval = (3.79, 6.21)

Necessary recorder points = 3.79 and 6.21

Question 3

Assuming that demand is normally distributed with a mean of 18 units and a standard deviation of 4 units, and lead-time is normally distributed with a mean of 5 days and a standard deviation of 2 days, what is the reorder point necessary to provide a 98% level of service?

Solution:

We are given

Mean = 18, SD = 4, confidence level = 98%

Recorder point for demand

X = mean + Z*SD

Critical value Z for 98% confidence level = 2.33

X = 18 + 2.33*4

X = 27.32

Recorder point for lead time

Mean = 5, SD = 2, confidence level = 98%

X = 5 + 2.33*2

X = 9.66

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